Не могу решить систему

x^2+y^2=8

x^3+2x^2y-xy^2-2y^3=0

Разделить 2 уравнение на y^3

(x/y) ^3+2 (x/y) ^2 - (x/y) - 2=0

x/y=t

t^3+2t^2-t-2=0

t^2 (t+2) - (t+2) = 0

(t^2-1) (t+2) = 0

(t-1) (t+1) (t+2) = 0

t=1; - 1; - 2

1) x^2+y^2=8

x/y=1

x=y

2y^2=8

y^2=4

y1=2

y2=-2

x1=2

x2=-2

2) x^2+y^2=8

x/y=-1

x=-y

y^2=4

y3=2

y4=-2

x3=-2

x4=2

3) x^2+y^2=8

x/y=2

x=-2y

4y^2+y^2=8

y^2=8/5

y5=sqrt (8/5)

y6=-sqrt (8/5)

x5=-2sqrt (8/5)

x6=2sqrt (8/5)

Ответ: (+-2; +-2), (2sqrt (8/5) ; - sqrt (8/5)), (-2sqrt (8/5) ; sqrt (8/5))